Question

A boy standing on a long railroad car throws a ball straight upwards. The car is moving on the horizontal road with an acceleration of $1{\text{ms}}^{-2}$ and the projection velocity in the vertical direction is $9.8\text{m/s}$. How far (in m) behind the boy will the ball fall? Take $g=9.8\frac{m}{s^2}$

Answer 1

Let the velocity of the car be $u$ when ball is thrown.
So, initial velocity of car = Horizontal velocity of ball.
$\therefore$ In horizontal direction, distance travelled by ball $S_B=ut$
Distance travelled by car is $S_C=ut+\frac{1}{2}a{t}^2$, where $t$ = time of flight of ball in the air

So, car has travelled extra distance of $S_C-S_B=\frac{1}{2}a{t}^2$
$\therefore t=\frac{2u}{g}=\frac{2\times 9.8}{9.8}=2\text{s},$ and hence, $S_C-S_B=\frac{1}{2}\times 1\times 2^2=2m$
So, the ball will drop $2\text{m}$ behind the boy